![y=x+ \frac{4}{x}](https://tex.z-dn.net/?f=y%3Dx%2B+%5Cfrac%7B4%7D%7Bx%7D+)
Найти наибольшее и наименьшее значение функциии
на отрезке
![[1,4]](https://tex.z-dn.net/?f=%5B1%2C4%5D)
Вычислим значение функции в критических точках:
![f'=(x+ \frac{4}{x})'=x'+ \frac{4'\cdot x-4\cdot x'}{x^2}=1- \frac{4}{x^2}\\\\ 1- \frac{4}{x^2}=0\\\\ \frac{x^2-4}{x^2}=0\\\\ x\neq0\\\\ x^2-4=0\\ x^2=4\\ x=\pm2\\\\ f(2)=2+ \frac{4}{2}=4\\\\ f(1)=1+4=5\\\\ f(4)=4+1=5](https://tex.z-dn.net/?f=f%27%3D%28x%2B+%5Cfrac%7B4%7D%7Bx%7D%29%27%3Dx%27%2B+%5Cfrac%7B4%27%5Ccdot+x-4%5Ccdot+x%27%7D%7Bx%5E2%7D%3D1-+%5Cfrac%7B4%7D%7Bx%5E2%7D%5C%5C%5C%5C%0A1-+%5Cfrac%7B4%7D%7Bx%5E2%7D%3D0%5C%5C%5C%5C%0A+%5Cfrac%7Bx%5E2-4%7D%7Bx%5E2%7D%3D0%5C%5C%5C%5C%0Ax%5Cneq0%5C%5C%5C%5C%0Ax%5E2-4%3D0%5C%5C%0Ax%5E2%3D4%5C%5C%0Ax%3D%5Cpm2%5C%5C%5C%5C%0Af%282%29%3D2%2B+%5Cfrac%7B4%7D%7B2%7D%3D4%5C%5C%5C%5C%0Af%281%29%3D1%2B4%3D5%5C%5C%5C%5C%0Af%284%29%3D4%2B1%3D5+++++)
Ответ:
![\max [-1;4] \ f(x)=f(1)=f(4)=5\\ \min [-1;4] \ f(x)=f(2)=4](https://tex.z-dn.net/?f=%5Cmax+%5B-1%3B4%5D+%5C+f%28x%29%3Df%281%29%3Df%284%29%3D5%5C%5C%0A%5Cmin+%5B-1%3B4%5D+%5C+f%28x%29%3Df%282%29%3D4)
Всегда принадлежат три точки -1 0 1
{6х+y=5Ⅰ×3
{2x-3y=-5
{18х+3у=15
{2х-3у=-5
{20х=10
{2х-3у=-5
{х=0,5
{2×0,5-3у=-5
{х=0,5
{-3у=-5-1
{х=0,5
{-3у=-6
{х=0,5
{у=2
Ответ:(0,5;2)
<span>А)
2m(a-b)-3n(b-a) = </span>2m(a-b)+3n(a-b) = (a-b)(2m+3n)
<span>Б)
18ab</span>³+27a²-45ab⁴ = 9a(2b³+3a-5b⁴)<span>
В)
a(x-y)+b(x-y)-c(y-x) = </span>a(x-y)+b(x-y)+c(x-y) = (x-y)(a+b+c)<span>
Г)
m(a-1)-2n+2an = </span>m(a-1)+(2an-2n) = m(a-1)+2n(a-1) = (a-1)(m+2n) <span>
Д)
32 ac</span>²+15cx²-10c³-48ax² = (32 ac²-48ax²)-(10c³-15cx²) =
= 16a(2c²-3x²)-5c(2c²-3x²) = (2c²-3x²)(16a-5c)